$x^2 + y^2+xy = 1$ , then find the minimum of $x^3 y + xy^3 +4$ x and y belongs to real numbers. $ x^2 + y^2+xy = 1 $.  then find the minimum value of $x^3 y + xy^3 +4$.
I assume  $ x = r \sin (w)$ and $ y = r\cos(w) $. $  x^3 y + xy^3 +4 = L $  which give me $ \frac{2}{3} \le r^2 \le 2 $ I am stuck after that.Its my  Humble request to help me after that.
 A: Hint: $x^2+y^2=1-xy$, so that $x^3y+xy^3+4=xy(x^2+y^2)+4=xy(1-xy)+4$. Substitute $z=xy$. What do you end up with?
A: Lagrangian multiplier for constrained function 
$$L(x,y)= F(x,y)-\lambda G(x,y) $$
set in the order given, partially differentiating with respect to both variables
$$ \dfrac{F_x}{F_y}=\dfrac{G_x}{G_y}= \lambda $$
$$\dfrac{2x+y}{2y+x}=\dfrac{3x^2y+x^3}{x^3+3y^2} $$
Cross multiplying to simplify we get two conditions
$$ (x^2-y^2)(x^2+y^2-xy)=0 $$
Minimal value verifiable by sign of second partial derivatives of $L(x,y)$..
A: Your method of solution is good. You just need to continue. By setting $x=r\cos\theta$ and $y=r\sin\theta$, we get$$\cos\theta\sin\theta=\frac1{r^2}$$ and so $\tfrac23\leqslant r^2\leqslant2$, as you found. We have $$x^3y+xy^3+4=r^2-r^4+4=\tfrac{17}4-(r^2-\tfrac12)^2,$$which is minimum when $|r^2-\frac12|$ is maximum, namely when $r^2=2$.
A: For $x=1$ and $y=-1$ we'll get a value $2$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$x^3y+y^3x+4\geq2$$ or
$$x^3y+y^3x+2(x^2+xy+y^2)^2\geq0$$ or
$$2x^4+5x^3y+6x^2y^2+5xy^3+2y^4\geq0$$ or
$$2x^4+4x^3y+2x^2y^2+x^3y+2x^2y^2+xy^3+2x^2y^2+4xy^3+2y^4\geq0$$ or
$$(x+y)^2(2x^2+xy+2y^2)\geq0.$$
Done!
